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http://jcm.sagepub.com/Materials

Journal of Composite

http://jcm.sagepub.com/content/16/4/268The online version of this article can be found at:

DOI: 10.1177/002199838201600402

1982 16: 268Journal of Composite MaterialsChoon T. Chon

Analysis of Tubular Lap Joint in Torsion

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268

Analysis of Tubular Lap Joint in Torsion

CHOON T. CHON

Engineering and Research StaffFord Motor Company

Dearborn, Michigan 48121

(Received January 18, 1982)

ABSTRACT

The stress distribution of the tubular lap joint whose adherends are com-posite materials is analysed with a closed form solution. In the analysis thestress components, rz and r, which are identically zero if both adherendsare isotropic materials, are introduced to satisfy the equilibrium equationsand the compatibility conditions in the region of lap joint. The stress concen-trations at and near the end of the joint as functions of the variousparameters, such as wrap angles, overlap length, and thickness of theadhesive layer are studied.

INTRODUCTION

THE EARLIEST THEORETICAL analysis of the (single) lap joint is that due toVolkersen [ 1 ] . In his analysis he ignores the tearing stresses resultingfrom the bending of the members and confines his attention to the determina-tion of the shearing stress distribution in the adhesive layer. A subsequent andnotable contribution to the analysis of this problem, due to Goland andReissner [2], shows that the bending of the adherend sheets caused by the ec-centricity of the applied tension produces a significant, direct stress perpen-dicular to the place of the adhesive layer, particularly at the edge of the joint,where its effect is to tear the sheets apart. Goland and Reissner proposed twoapproximation theories in their paper. In the first, it is assumed that theadhesive layer is so thin that its effect on the flexibility of the joint is negligi-ble. In the second, which has been used more often since, it is assumed that theflexibility of the joint arises mainly from that of the adhesive. In this secondapproximation the adhesive layers are treated as a system of infinitesimal spr-ings placed between the two plates. The choice of the appropriate methoddepends on the elastic and geometric properties of the joint under considera-tion, but in many practical cases the basis for making this choice is not clear.

For the tubular lap joint analysis, Lubkin and Reissner [3] considered thedistribution of stress in the adhesive lap joint between thin cylindrical tubes of

J. COMPOSITE MATERIALS, Vol. 16 (July 1982), p. 268

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circular cross-section for the case in which the tubes are axially loaded. Alwarand Nagaraja [4] considered the same problem, but they treated the adhesivematerial as viscoelastic. For the tubular lap joint under torsional loading,Volkersen [5] has given a closed form solution for the shear stresses. Stresses inthe tubular lap joint under torsional loading have also been investigated byAdams and Peppiatt [6]. Stresses in tubular lap joints under other loading con-ditions, such as external and internal pressures, have been investigated byTerekhova and Skoryi [7], who neglected the effect of adherend bending. Allof them, however, considered only cases where the adherends were isotropicmaterials.

Torsion of a Laminated Tube

We consider a composite shaft under a twisting moment T in cylindricalcoordinates (1., z, ~) as shown in Figure 1. The helical angle 0 is measured fromz-axis as shown in Figure 1. The cross-section consists of N helically woundcomposite layers (Figure 2).

In order to derive the constitutive rela-

tions, the following assumptions arenecessary:

(i) The laminate is macroscopicallyhomogeneous, linearly elastic, and initiallystress free,

(ii) the fibers are homogeneous, linearlyelastic, isotropic, regularly spaced andperfectly aligned,

(iii) the matrix is homogeneous, linearlyelastic and isotropic, and in addition,

(iv) no voids exist in the fibers or matrixor between them (i.e., the bonds between thefibers and matrix are perfect)We first consider a special case when all

fibers are parallel to the ,z-axis (9 = 0 ),which can be assumed to be macroscopicallytransversely isotropic. Then, the com-ponents of the compliance matrix can be ex-pressed in terms of the properties of thefibers, matrix material and the relativevolumes of fibers and matrix:

Figure 1. Coordinate system of atube (r, +, z), the angle denotes the

helical angle.

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270

Figure 2. Geometry of the cross-section of the tube.

otherwise Sij = 0.The details are derived in Reference [8]. Therefore, the compliance matrix

for any helical angle 0 of each layer can be obtained by performing the simpletensor transformation.

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or

where

It should be noted that

where 6ik is a Kronecker delta function.Since the cross-section is circular, we may assume that stresses are indepen-

dent of the coordinate angle ~. Therefore, the equilibrium equations in eachlayer become

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If we also assume that displacements are independent of , then, in eachlayer,

It should be noted that in equations (1)-(5), the superscript k(=1,---, l~ whichdenotes the number of layers is omitted for convenience.Now we imagine an engine shaft subjected to torsional loading: a torque

about the axis of the shaft is produced at one end by the driving mechanismand at the other end by an opposite torque that originates in the mechanismthat receives the power. In steady motion both torques have the samemagnitude. For a straight driveshaft, a torque acts about the shaft axis at oneend, and whether the other end is fixed or subject to the opposite momentmakes no difference. If we consider any cross-section in the central region of along cylinder, the boundary conditions should be in the following form:

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In addition, we have the continuity conditions:

where

We seek a solution of the equations with the given boundary conditions(equations (1)-(9)) by means of St. Venant semi-inverse method. That is, wepropose certain formulations for some of the variables and then justify theseformulations and find the solutions of the remaining variables by solving thefield equations. The final solutions of equation (4) are given from [9] asfollows:

where

If

we replace

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by

Also,

Using the boundary conditions in equations (6)-(9) and the continuity condi-tions equation (10), (4N+ 2) coefficients .4~, ~, C

On the basis of these assumptions, we can derive new field equations for thejoint portion. The system considered is shown schematically in Figure 3, inwhich the thickness of the adhesive layer, 1}, is exaggerated for clarity. Thejoint length, the outer radius of the inner tube and the inner radius of the outertube are denoted by 2c, rIo and rni respectively.

Equilibrlum Equations - When the lap joint is subjected to a twisting mo-ment, T, twisting moments Ti, 7n, resultant forces Fi, Fn and resultantpressures pi, pn are produced in both adherends at any given cross section.Then from the global equilibrium conditions, we should have

It should be noted that FI, Fn, pI and j?n are identically zero for the case ofisotropic adherends. If we denote the stresses in the adhesive by 0, T ;, and Tas shown in Figure 4 balancing the twisting moments, the forces and thepressures acting on the elements over a length 4iz, we obtain additionalequilibrium equations, such as

and

where

Figure 4. Definitions of stresses and strains on the adhesive layers.

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276

Constitutive Equations

(1) Adhesive layer - We consider the adhesive layer to be isotropic.Therefore, the constitutive equations become

where Et , yn, and y~ are strains in the adhesive shown in Figure 4.(2) Adherends - We use equation (2) as the constitutive equations for both

adherends but with different S,~s; viz., we use S and 5~ for the outer layer ofthe inner tube and the inner layer of the outer tube, respectively.

Compatibility Conditions - It can be easily shown from equation (5) thatthe compatibility conditions among strains (E#, y~, y), (c~, t~, e~,j, and (&dquo;, IIi ~IIi y$)I) in an element of the joint of length ~.z are

and

However, since F-, in the tube is constant (see equations (5 & 13)),

Boundary Conditions - The boundary conditions at the joint edges aregiven by expressions of the type,

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277

which also satisfy the global equilibrium conditions (equation (14)). It shouldbe noted that the conditions, Fi = FiI = 0, are from the edge conditions of thelap joint for the shear stress component, namely, Il ( t c) = 0 (see equation(16)).

Analysts of the Stress Distribution - In this section, we solve for thetwisting moments (Ti and Tn), axial forces (FI and Flo, and pressures (pi andpI~. Since combined loadings can be handled by superposition, the totalstrains are decomposed into three parts; i.e., strains due to a twisting moment,lj, strains due to an axial force, ~, and strains due to a pressure, ~, or

It should be noted that the strains, E and may be obtained by using ap-propriate boundary conditions from equation (6) through equation (9);

for

for the pressure from inside, or

for the pressure from outside of the tube. Otherwise, all other boundary condi-tions in equations (6-9) are identically zero. Differentiating equations (15)and (16) with respect to z and rearranging equations (2), (17), (18), (19), (20),(21), and (23), we obtain

and

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Eliminating Fn and Pn from equations (24), (25), and (26) gives

where C, and Ci are constants to be found later, and

are strains due to unit load (unit torque, unit axial force, or unit pressure).The solution of equation (27), is

where

Coefficients Ci (i =1, ... ,6) in equation (28) can be found from boundaryconditions (equation (22)). Then, Pn, Pn, Or, T~ and T~ can be computedsubsequently from appropriate equations.

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279

Numerical Results and Discussions

Table 1.

Figure 5. Stress distribution along the lapjoint for 0 (or 90) glass-steel as adherends.Also shown is the stress distribution for steel-

steel lap joint.

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( b)

Figure 6. Stress distribution for -45 and for + 45 0 glass-steel lap joint.

Flaurt 7. Maximum stresses T~., T~ and o# vs. helical angle. Dotted lines are foruni-directional and solid lines are for angle ply shaft. Both are for glass-steel lap

joint.

In our numerical calculations, unless otherwise specified, we consider acomposite tube, in which glass fiber is bonded to the steel tube. Themechanical properties of fiber and matrix materials and dimensions of thetube used in the actual calculations in this section are listed in Table 1. Weshow the distribution of the normal stress, o~, and the shear stresses T~ and T~,developed in the lap joint in Figure 5 and 6 for unidirectional and angle plycomposite tubes. In Figure 5, we also plot the distribution of shear stress when

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Figure 8. Stress distribution along the lap Figure 9. The same as in Figure 8 except forjoint for 2cltc being 1, 10, and 30. They are 8 = 0 (or 90).for the case of -45 uni-directional tube.

both adherends are made of steel. It is interesting to note that when the windingangle is 0 (or 90 ), the normal stress, va, and the shear stress, T~ become iden-tically zero, but a large stress concentration in T~ occurs at the edge of thejoint. (When both adherends are steel the maximum stress concentration is on-ly about 1/10 of this). The large stress concentration is mainly due to an&dquo;adherend stiffness imbalance&dquo;. As stated by Hart-Smith [12], &dquo;... whenthere is a difference in this stiffness of the two adherends joined with anadhesive, the shear strains are intensified at the end of the joint from which theless stiff adherend extends.&dquo; In Figure 6, we show that stress concentrations inT~ and o/&dquo; occur at the edge of the joint for a tube in which all of the fiber iswrapped either -45 or 45

helical angle. In addition, due to the free edgeboundary condition T~ becomes maximum near the edge of the joint. It shouldbe emphasized that angle ply construction significantly reduces the stress con-centrations of T~ and v; . In Figure 6, the normal stress Or for t 45 is not

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282

Figure 10. Maximum stress vs. overlap length 2cltc. This~gure showsa region of constant stress.

Figure 11. Maximum stmma at the end of the lap joint vs. thicknessof the adhesive layer, n.

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shown because of its negligibly small magnitude. The maximum stresses of Tn,~, and o; are plotted as functions of wrap angles in Figure 7. The solid linesrepresent the angle ply shaft and the dotted lines represent the unidirectionals...